1965 AHSME Problems/Problem 2

A regular hexagon is inscribed in a circle. The ratio of the length of a side of the hexagon to the length of the shorter of the arcs intercepted by the side, is:

$\textbf{(A)}\ 1: 1 \qquad \textbf{(B) }\ 1: 6 \qquad \textbf{(C) }\ 1: \pi \qquad \textbf{(D) }\ 3: \pi \qquad \textbf{(E) }\ 6:\pi$

Solution

Suppose that each side of the hexagon is $3$ . Then the distance from each vertex of the hexagon to the center is also $3$ , so that the circle has radius $3$ . Since the circle has circumference $2\pi(3) = 6\pi$ , the arc intercepted by any side (which measures $60^\circ$ ) has length $\frac{1}{6}*6\pi = \pi$ , and we are done.

1965 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1965 AHSME Problems/Problem 2

Solution

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